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Lorico [155]
3 years ago
9

Which interval contains a local minimum for the grafted function? [-4, -2.5] [-2, -1] [1, 2] [2.5, 4]

Mathematics
1 answer:
olga_2 [115]3 years ago
7 0

Answer:

Last option [2.5, 4]

Step-by-step explanation:

The local minima of a function are the points where the slope of the curve is zero and the function reaches a minimum value.

If, on the contrary, the function reaches a maximum value, then that point is a local maximum.

Observe, for example, the point (3, -4). Note that a line tangent to that point would be horizontal, that is, of slope m = 0. This point is a minimum of the function, because there are no values x to the right x = 3 or to the left of x = 3 for which f(x)

Now we must search among the options that interval contains at this point.

The intervals [-4, -2.5] [-2, -1] do not contain any local minimum

The intervalor [1, 2] contains a local maximum.

The only interval that contains a local minimum is [2.5, 4], which contains the point (3, -4)

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A coin is tossed 71 times and 34 heads are observed. Is the coin fair? Use a 98% level confidence interval to base your inferenc
Fofino [41]

Answer:

a) \hat p=\frac{34}{71}=0.48 estimated proportion of heads observed.

b)  Se= \sqrt{\frac{0.479(1-0.479)}{71}}=0.0593

c) z_{\alpha/2}=2.33

d) 0.479 - 2.33 \sqrt{\frac{0.479(1-0.479)}{71}}=0.341

e) 0.479 + 2.33 \sqrt{\frac{0.479(1-0.479)}{71}}=0.617

f) In order to be fair the coin needs to have included the 0.5 on the interval, and for this case it's included, so we can say that at 2% of significance we can say that the coin is fair.

Step-by-step explanation:

Notation and definitions

X=34 number of heads observed.

n=71 random sample taken

\hat p=\frac{34}{71}=0.479 estimated proportion of heads observed.

p true population proportion of heads observed.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})

The point estimator for the population proportion of heads is: Answer with two decimal precision.

\hat p=\frac{34}{71}=0.48 estimated proportion of heads observed.

The standard error in this estimate is: Answer with four decimal precision.

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

And the standard error is given by:

Se= \sqrt{\frac{\hat p(1-\hat p)}{n}}

Se= \sqrt{\frac{0.479(1-0.479)}{71}}=0.0593

The correct table value for the 98% level confidence interval is: Answer with two decimal precision

For the 98% confidence interval the value of \alpha=1-0.98=0.02 and \alpha/2=0.01, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.33

The lower end point of the CI is: Answer with three decimal precision.

And replacing into the confidence interval formula we got:

0.479 - 2.33 \sqrt{\frac{0.479(1-0.479)}{71}}=0.341

The upper end point of the CI is: Answer with three decimal precision.

0.479 + 2.33 \sqrt{\frac{0.479(1-0.479)}{71}}=0.617

And the 90% confidence interval would be given (0.341;0.617).

Based on the confidence interval, it is plausible that the coin is fair.

In order to be fair the coin needs to have included the 0.5 on the interval, and for this case it's included, so we can say that at 2% of significance we can say that the coin is fair.

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